If $\displaystyle \tan \mathrm{A}=\sqrt{3}$; where A is an acute angle, then find the value of $\displaystyle \frac{\sin ^{2} \mathrm{~A}}{1+\cos ^{2} \mathrm{~A}}$.

\(\displaystyle \tan \mathrm{A}=\sqrt{3}=\tan 60^{\circ}\) \[\begin{aligned} & \Rightarrow \mathrm{A}=60^{\circ} \\ & \frac{\sin ^{2} \mathrm{~A}}{1+\cos ^{2} \mathrm{~A}}=\frac{\sin ^{2} 60^{\circ}}{1+\cos ^{2} 60^{\circ}} \\ & =\frac{\left(\dfrac{\sqrt{3}}{2}\right)^{2}}{1+\left(\dfrac{1}{2}\right)^{2}} \\ & =\frac{3}{5} \end{aligned} \]